Let $z_1, z_2$ be two complex numbers such that $\bar{z}_1-i \bar{z}_2=0$ and $\arg \left(z_1 z_2\right)=\frac{3 \pi}{4}$, then $\arg \left(z_1\right)=$

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Aicharya · Accepted Answer

Given $\bar{z}_1-i \bar{z}_2=0$ $\begin{array}{ll} \Rightarrow & \bar{z}_1=i \bar{z}_2 \ \Rightarrow & \bar{z}_1=\overline{i z_2} \ \Rightarrow & z_1=-i z_2 \end{array}$ Clearly argument of $z_1=$ argument of $z_2-\frac{\pi}{2}$ or argument $z_1=$ argument $z_2-\frac{\pi}{2}$ $\Rightarrow$ argument $z_2=\operatorname{argument} z_1+\frac{\pi}{2}$ Let argument $z_1=\alpha$ Then, given argument $\left(z_1 z_2\right)=\frac{3 \pi}{4}$ $\Rightarrow$ argument $z_1+\operatorname{argument} z_2=\frac{3 \pi}{4}$ $\begin{aligned} \alpha+\alpha+\frac{\pi}{2} & =\frac{3 \pi}{4} \ 2 \alpha & =\frac{3 \pi}{4}-\frac{\pi}{2} \Rightarrow \alpha=\frac{\pi}{8} \end{aligned}$

Let \(z_1, z_2\) be two complex numbers such that \(\bar{z}_1-i \bar{z}_2=0\) and \(\arg \left(z_1…

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